Question

Which of the following lines is perpendicular to \(\mathrm{x}+2 \mathrm{y}+3=0\) ? (1) \(2 \sqrt{2} \mathrm{x}-\mathrm{y}+3=0\) (2) \(\sqrt{2} \mathrm{x}+\sqrt{2} \mathrm{y}-5=0\) (3) \(2 \sqrt{2} \mathrm{x}-\sqrt{2} \mathrm{y}+3=0\) (4) \(x+\sqrt{2} y+4=0\)

Short answer

a) \(2\sqrt{2}x-y+3=0\) b) \(\sqrt{2}x+\sqrt{2}y-5=0\) c) \(2\sqrt{2}x-\sqrt{2}y+3=0\) d) \(x+\sqrt{2}y+4=0\) Answer: c) \(2\sqrt{2}x-\sqrt{2}y+3=0\)

Step-by-step solution

Find the slope of the given line

Rearrange the equation \(x+2y+3=0\) to find the slope. First, move the terms involving \(x\) and \(y\) to the other side as follows: \(y = -\frac{1}{2}x - \frac{3}{2}\). In this form, the slope of the given line is \(-\frac{1}{2}\).

Calculate the slope of option (1)

Rearrange the equation \(2\sqrt{2}x-y+3=0\) to a slope-intercept form: \(y=2\sqrt{2}x+3\). The slope of option (1) is \(2\sqrt{2}\).

Calculate the slope of option (2)

Rearrange the equation \(\sqrt{2}x+\sqrt{2}y-5=0\) to the slope-intercept form: \(y=-x+5/\sqrt{2}\). The slope of option (2) is \(-1\).

Calculate the slope of option (3)

Rearrange the equation \(2\sqrt{2}x-\sqrt{2}y+3=0\) to the slope-intercept form: \(y=2x+3/\sqrt{2}\). The slope of option (3) is \(2\).

Calculate the slope of option (4)

Rearrange the equation \(x+\sqrt{2}y+4=0\) to the slope-intercept form: \(y=-\frac{1}{\sqrt{2}}x-\frac{4}{\sqrt{2}}\). The slope of option (4) is \(-\frac{1}{\sqrt{2}}\).

Identify the line with a slope that gives a product of -1 with the given line

Compare the slopes of each option to the slope of the given line (\(-\frac{1}{2}\)). The lines are perpendicular if the product of their slopes is -1. (1) \(-\frac{1}{2} * 2\sqrt{2} = -\sqrt{2}\) (2) \(-\frac{1}{2} * -1 = \frac{1}{2}\) (3) \(-\frac{1}{2} * 2 = -1\) (4) \(-\frac{1}{2} * -\frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}\) Option (3) has a slope that when multiplied by the given line's slope results in -1. Therefore, the line \(2\sqrt{2}x-\sqrt{2}y+3=0\) (option (3)) is perpendicular to the given line \(x+2y+3=0\).

Key concepts

Slope of a Line

Understanding the slope of a line is critical when studying geometry and algebra. It represents how steep a line is and the direction it's going. Imagine you're hiking: the slope tells you how quickly the elevation changes as you walk forward.

To find a line’s slope, we often rearrange the equation into the slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. It shows how much y changes for a one-unit change in x. For example, a slope of 2 means that for every step you take horizontally, you climb 2 units vertically.

In the context of perpendicular lines, slopes have a special relationship. If one line has a slope of a, the line perpendicular to it will have a slope of -1/a if a is not zero. This inverse relationship is essential for determining perpendicularity between two lines.

Slope-Intercept Form

The slope-intercept form of a line, y = mx + b, is a straight shooter when it comes to identifying the slope and y-intercept of a line. It's your algebraic GPS for navigating equations. In this form, m stands for the slope, and b identifies the point where the line crosses the y-axis.

For instance, if we manipulate the equation x + 2y + 3 = 0 into the slope-intercept form, we would get y = -1/2x - 3/2. This way, it's crystal clear that the slope is -1/2 and the line crosses the y-axis at -1.5. When working with perpendicular lines, transforming equations into this form simplifies the process of finding slopes, making it easier to spot that crucial perpendicular relationship.

Product of Slopes

The product of slopes becomes a spotlight when it shines on the relationship between two lines. Particularly, when you're on the lookout for perpendicular lines, you need the product of their slopes to be -1. This fact is like a secret handshake for perpendicular lines; only they know it.

Take our earlier slope example: if one line has a slope of -1/2, and you're searching for a line perpendicular to it, the slope of this new line should satisfy the equation (-1/2) * m = -1, where m is the slope of the perpendicular line. We solve for m to find it equals 2. Therefore, if another line has a slope of 2, these two are the perpendicular pals we seek. It's a simple check that's both efficient and reliable for verifying perpendicularity in the geometric landscape.